Problem: $g(x) = -x^{3}+6x^{2}-x-6+4(f(x))$ $f(n) = -3n+2$ $ g(f(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = (-3)(1)+2$ $f(1) = -1$ Now we know that $f(1) = -1$ . Let's solve for $g(f(1))$ , which is $g(-1)$ $g(-1) = -(-1)^{3}+6(-1)^{2}-(-1)-6+4(f(-1))$ To solve for the value of $g$ , we need to solve for the value of $f(-1)$ $f(-1) = (-3)(-1)+2$ $f(-1) = 5$ That means $g(-1) = -(-1)^{3}+6(-1)^{2}-(-1)-6+(4)(5)$ $g(-1) = 22$